Question

A mixture of $N_2$ and $H_2$ is the molar ratio $1:3$ at 50 atm and $650$ is allowed to react till equilibrium is obtained. The $N{H}_3$ present at equilibrium is 25% by weight, $K_p$ for,
$N_2+3H_2\rightleftharpoons 2N{H}_3$

is $X\times {10}^{-4}{atm}^{-2}$. $\left(1000X\right)$ is_________.

• A. 1434
• B. 35844
• C. 21370
• D. none of these

Answer 1

Answer: (C)

21370

	$N_2$	$H_2$	$N{H}_3$
Initial number of moles	1	3	0
Equilibrium number of moles	$(1-x)$	$3(1-x)$	$2(1-x)$

Ammonia present at equilibrium is 25% by weight.
$\frac{2(1-x)\times 17}{28+6}=\frac{1}{4}$
$1-x=0.25$
$x=0.75$

	$N_2$	$H_2$	$N{H}_3$
Equilibrium number of moles	0.25	0.75	0.5
Equilibrium mole fraction	0.167	0.5	0.33

	$N_2$	$H_2$	$N{H}_3$
Equilibrium partial pressure	0.167	25	16.7

$K_p=\frac{(16.7{)}^2}{8.35(25{)}^3}=21.37\times {10}^{-4}=X\times {10}^{-4}$
$X=21.37$
$1000X=21370$